Question: Consider the parametric curve: $\begin{aligned} x&=\dfrac{4}{t} \\\\ y&=\ln(-t) \end{aligned}$ Which integral gives the arc length of the curve over the interval from $t=-12$ to $t=-5$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\int_{-12}^{-5} \sqrt{\dfrac{16}{t^4}-\dfrac{1}{t^2}}\,dt$ (Choice B) B $\int_{-12}^{-5} \sqrt{-\dfrac{4}{t^2}+\dfrac{1}{t}}\,dt$ (Choice C) C $\int_{-12}^{-5} \sqrt{-\dfrac{4}{t^2}-\dfrac{1}{t}}\,dt$ (Choice D) D $\int_{-12}^{-5} \sqrt{\dfrac{16}{t^4}+\dfrac{1}{t^2}}\,dt$
Explanation: This is the formula for the arc length of a parametric curve over the interval $[a, b]$ : $\int_a^b\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\, dt$ [Where does this formula come from?] Let's find $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ : $\begin{aligned} \dfrac{dx}{dt}&=\dfrac{d}{dt}\left[\dfrac{4}{t}\right] \\\\ &=-\dfrac{4}{t^2} \\\\\\ \dfrac{dy}{dt}&=\dfrac{d}{dt}\left[\ln(-t)\right] \\\\ &=\dfrac{-1}{-t} \\\\ &=\dfrac{1}{t} \end{aligned}$ Now we can find the expression for the arc length: $\begin{aligned} &\phantom{=}\int_{-12}^{-5} \sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt \\\\ &=\int_{-12}^{-5} \sqrt{\left(-\dfrac{4}{t^2}\right)^2+\left(\dfrac{1}{t}\right)^2}\,dt \\\\ &=\int_{-12}^{-5} \sqrt{\dfrac{16}{t^4}+\dfrac{1}{t^2}}\,dt \end{aligned}$ In conclusion, this integral gives the arc length of the curve over the interval from $t=-12$ to $t=-5$ : $\int_{-12}^{-5} \sqrt{\dfrac{16}{t^4}+\dfrac{1}{t^2}}\,dt$